The problem this and the other replies miss is that the standard definition of division is multiplication by the inverse. The entire argument rests on a notational slight of hand. The property that held before -- that _when defined_ division has the inverse property -- no longer holds. Thus many equational identities that otherwise would hold do not hold.

 > The problem this and the other replies miss is that the standard definition of division is multiplication by the inverse.Try to state this definition formally. The statement: ∀ x,y . x/y = xy⁻¹ is not a theorem of fields or a definition of division. However, ∀ x, y . y ≠ 0 ⇒ x/y = xy⁻¹ is, but is completely unaffected by defining division at 0. Those who think they see a problem rely on informal and imprecise definitions. Could you formally state a theorem that is affected? That would help you get around issues that are merely artifacts of imprecision.But let's entertain you, and state that what we really mean by the informal and vague statement, "division is the inverse of multiplication," could be stated formally as: ∀ x ∈ dom(1/t). x(1/x) = 1 You are absolutely correct that this equational theorem is broken by extending the domain of division. However, there is absolutely no way to say that the formalization of this theorem isn't actually ∀ x ≠ 0 . x(1/x) = 1 because the two are equivalent. You cannot then claim that something is necessarily broken if you choose to pick a formalization that is indeed broken, while an equivalent formalization exists, that is not broken (not to mention that the formalization that is broken requires a strictly richer language). All that means is that your formalization in this case is brittle, not that laws are broken.
 I have to disagree -- this isn't sleight of hand. The standard definition isn't being violated here, because standard division isn't a total function. The denominator's domain in Hillel's function is a proper superset of the standard domain: when restricted to the standard domain, the two functions are precisely equivalent. Therefore, every standard identity still holds under Hillel.The hole that he is filling here isn't one that he bored into the standard definition, but a hole that the standard definition already admitted. If something is explicitly undefined, there's nothing mathematically wrong with defining it, as long as the definition doesn't lead to inconsistency.
 > Onlookers reading these comments probably think those of us harping on this point are anal pedants with a mathematical stick up our ass. But this thread is increasingly illustrating my central point, which is that the author shouldn't have tried to justify numerical operation definitions in a programming language using field axioms of all things.You can't use your own stubbornness to justify itself.I'm waiting for you to justify your claim that this extension to division breaks any of the field axioms. pron even made you a nice list of them.Just name one equation/theorem that the new axiom would break.I'm completely open to being convinced! But so far you've only given arguments about giving a multiplicative inverse to zero. Everyone agrees on that. It's the wrong argument.
 Because a divisor cannot exist unless it is a multiplicative inverse. Therefore 0 is not a divisor.This is getting to be Kafkaesque...it breaks the field axioms themselves. How many different explanations and external resources do I need to provide in this thread for you to be convinced that this is not a controversial point in modern mathematics? I just explained it in the comment you responded to.You have exactly two options here.If you define x/0, that definition must interact with the rest of the definitions and elements of the field. To maintain multiplicative closure (a field axiom!) there must be a unique y element equal to x/0. So tell me how you will define x/0 such that x is not equal to the product of 0 and y. Regardless of what you think the author has shown, the burden of proof is not on me at this point to show that you can't do it, because it follows directly from the field axioms. Trying to impose a one-off bizarro divisor function defined only on {0} is not only mathematically inelegant, it immediately eliminates the uniqueness of all field elements. Therefore your "field" just becomes {0}, and since it lacks a multiplicative identity it ceases to be a field. There is your contradiction. Why don't you tell me how you're going to prove any equation defined over a field that relies on the uniqueness or cancellation properties of fields?On the other hand, let's say you tell me you want define x/0 so that the definition doesn't interact with any of the field definitions or elements. Then you haven't actually introduced any new operation or definition, you've just designed a notation that looks a lot like division but has no mathematical bearing on the field itself (i.e. absolutely nothing changes, including for 0). That's not a divisor function, it's just a confusing shorthand. You can't just add another axiom to a field and call it a field.If you believe I'm stubborn, that's fine. I might be a poor teacher! There are ample resources online which will patiently explain this concept in mind numbing detail. It boggles my mind that there are people in this thread still fighting an idea in earnest which has been settled for over a century.
 > Because a divisor cannot exist unless it is a multiplicative inverse. Therefore 0 is not a divisor.So there are two separate issues here. One is whether we can extend the definition of the "/" operator, and the other is whether we call it "division".I'm not interested in what we call it. I'm interested in the claim that extending "/" will break the field.The dichotomy you're talking about is wrong. The two options are not "multiplicative inverse" and "does not interact with anything". "1/0 = 0" interacts with plenty! If I make a system where it's an axiom, I can calculate things like "1/0 + 5" or "sqrt(1/0)" or "7/0 + x = 7". I can't use it to cancel out a 0, but I can do a lot with it.> It boggles my mind that there are people in this thread still fighting an idea in earnest which has been settled for over a century.Remember, the question is not "should this be an axiom in 'normal' math?", the question is "does this actually conflict with the axioms of a field?"> You can't just add another axiom to a field and call it a field.Yes you can. There is an entire hierarchy of algebraic structures. Adding non-conflicting axioms to an X does not make it stop being an X.